我下载了波士顿的shapefile,并希望使用下面的代码将其绘制出来。但是它给了我一个错误ValueError: lat_0 must be between -90.000000 and 90.000000 degrees
结果coords
有值(33869.92130000144, 777617.2998000011, 330800.31099999696, 959741.1853)
为什么这么大?
Boston shapefile is obtained here
代码
# Import Boston shapefile
shapefilename = 'ZIPCODES_NT_POLY'
shp = fiona.open(shapefilename + '.shp')
coords = shp.bounds
shp.close()
w, h = coords[2] - coords[0], coords[3] - coords[1]
extra = 0.01
m = Basemap(
projection='tmerc', ellps='WGS84',
lon_0 = np.mean([coords[0], coords[2]]),
lat_0 = np.mean([coords[1], coords[3]]),
llcrnrlon = coords[0] - extra * w,
llcrnrlat = coords[1] - extra * h,
urcrnrlon = coords[2] + extra * w,
urcrnrlat = coords[3] + extra * h,
resolution = 'i', suppress_ticks = True)
错误
ValueError: lat_0 must be between -90.000000 and 90.000000 degrees
答案 0 :(得分:3)
您需要从Eastings和Northings的原生投影重新投影到另一个坐标参照系。如果您需要纬度和经度,通常为WGS84或EPSG:4326。以下是重新投影的方法:
import fiona
import pyproj
from functools import partial
from shapely.geometry import box
from shapely.ops import transform
shp = fiona.open('ZIPCODES_NT_POLY.shp', 'r')
p_in = pyproj.Proj(shp.crs)
bound_box = box(*shp.bounds)
shp.close()
p_out = pyproj.Proj({'init': 'EPSG:4326'}) # aka WGS84
project = partial(pyproj.transform, p_in, p_out)
bound_box_wgs84 = transform(project, bound_box)
print('native box: ' + str(bound_box))
print('WGS84 box: ' + str(bound_box_wgs84))
原生方框:POLYGON((330800.310999997 777617.2998000011,330800.310999997 959741.1853,33869.92130000144 959741.1853,33869.92130000144 777617.2998000011,330800.310999997 777617.2998000011))
WGS84包装盒:POLYGON((-69.93980848410942 41.23787282321487,-69.899038698261 42.87724537285449,-73.53324195423785 42.8704709990465,-73.48147096070339 41.2312695091688,-69.93980848410942 41.23787282321487))
否则,Basemap所需的大多数参数都在shp.crs
中(请查看)。
答案 1 :(得分:0)
必须更改坐标参考系。我推荐geopandas
:
import geopandas as gpd
import matplotlib.pyplot as plt
# load zips with the source projection
shapefilename = 'ZIPCODES_NT_POLY'
zips = gpd.read_file(shapefilename + '.shp')
# convert projection to familiar lat/lon
zips = zips.to_crs('epsg:4326')
zips.plot()
plt.show()